Answer to Question #228413 in Mechanical Engineering for Unknown346307

Question

The properties of a system, during a reversible constant pressure non￾flow process at p = 1.6 bar, changed from v1 = 0.3 m3/kg, T1 = 20°C to v2 = 0.55 m3/kg, T2 = 260°C.

The specific heat of the fluid is given by

cp = 1.5+74/(T+45)kJ/kg°c

 , where T is in °C.

Determine : (i) Heat added/kg ;

(ii) Work done/kg ;

(iii) Change in internal energy/kg ;

(iv) Change in enthalpy/kg

Answer 1

$\upsilon_1=0.3m^{3}/kg$

T₁=20°C

$\upsilon_2=0.55m^{3}/kg$

T₂=260°C

p=1.6 bar

Specific heat at constant pressure,c_p=$(1.5+\frac{74}{T+45})kJ/kg°C$

(i)heat added/kg :

$Q=\int_{T_1}^{T_2}c_pdT$

$=\int_{20}^{260}(1.5+\frac{74}{T+45})dT$

$=|1.5T+74log_e(T+45)|_{20}^{260}$

=1.5(260-20)+74$*log_e(\frac{260+45}{20+45})$

=474.398kJ

$\therefore heat \ added/kg=474.398kJ/kg$

(ii)Work done/kg :$W=\int_{\upsilon_1}^{\upsilon_2}pdv=p(\upsilon_2-\upsilon_1)$

=1.6*10⁵(0.55-0.3)N.m

=40*10³J=40kJ

$\therefore work\ done=40kJ/kg$

(iii)change in internal energy/kg : $∆u=Q-W$

=474.398-40=434.398 kJ/kg

(iv)change in enthalpy/kg (for non flow process):

∆h=Q

=474.398 kJ/kg